Just concentrate on what you have and start limiting the values of the digits step by step
The problem: 4th Hard number system question for CAT
The sum of thousands and tens digits in a 4 digit number is 4, the sum of hundreds and units digit is 15, and units digit is greater than the thousands digit by 7. Among all such numbers satisfying these conditions, find the number, the sum of product of whose thousands digit and units digit and hundreds digit and tens digit is the least.
Hint: For quick solution of this 4th hard number system question for CAT, limit the possible values of the digits by variable linking of the conditions.
Solution to the 4th hard number system question for CAT: Step by step limiting constraints on the digit values by variable value linking technique
First step: Problem definition in mathematical terms
Let $a$, $b$, $c$ and $d$ respectively be the thousands, hundreds, tens and units digits of the number $abcd$.
By the first part of the problem description,
One more independent linear equation would have been needed to solve for the values of all four digits. Understandably, that is not to be.
By the second part of the problem description,
Solution would be the number having least value of $(ad+bc)$ among all numbers satisfying the above three equations.
This completes translating the problem description into maths relations that we can analyze and manipulate for solving this problem quickly.
A look at the equations and the condition above and it is clear that,
The possible values of the digits are to be evaluated in terms of each other in as simple relations as possible, and more importantly,
Among the three equations, the equations (3) being most restrictive, it is the most important and to be taken up first.
Note: Solution speed would depend on how quickly you select the equation (3) as the starting point and proceed from there.
$d=a+7$, and as $a$ and $d$ both can have values ranging from 1 to maximum 9, possible values of the two are,
$d=8$, $a=1$ or $d=9$, $a=2$.
The two possible values of $d$ next puts restrictions on value of $b$ on the basis of equation (2),
$\Rightarrow$ Possible digit values,
$\Rightarrow$ $a=1$, $d=8$, and so, $b=7$ and,
$\Rightarrow$ $a=2$, $d=9$, and so, $b=6$.
Till now we have only two sets of possible values of the three digits. In each combination, the values are dependent on or bonded with each other through the equations.
This is a variable value linking process and the last of this process would give us the two possible values of $c$ from equation (1),
$\Rightarrow$ Possible digit values,
$\Rightarrow$ $b=7$, $d=8$, $a=1$, and so, $c=3$, ..........Combination 1, and,
$\Rightarrow$ $b=6$, $d=9$, $a=2$, and so, $c=2$. ..........Combination 2.
It takes little time to calculate, values of $ad+bc$ for the two combinations,
For Combination 1,
For Combination 2,
With lower value of $ad+bc$ between the two, Combination 1 gives us the number we want. It is,
$a=1$, $b=7$, $c=3$, $d=8$ $\Rightarrow$ $1738$.
No special concept has been used in the solution except taking best course of action to increase clarity into the problem step by step in fewest number of steps.
This is systematic problem solving that automatically produces correct result in optimum time.
The only problem solving technique that naturally occurred to us is the Variable value linking technique.
In solving reasoning puzzles, we had to use an abstract form of the technique, the linking bonded member structures.
Here, the digits are the members of the structure of each Combination in which they are bonded or fixed. These two bonded structures or Combinations are expanded step by step by linking with the other similar structures bonded in the equations.
In simple terminology, we have named this process as, Variable value linking technique which is good enough in this case.
Lastly we repeat, you would be able to quickly solve a problem that describes multiple conditions,
If you directly fix on the most important condition to start with. In this problem, it is the equation (3).
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